S2S 12. LARGER GROUPS OVER Fz IN .Cj(G, T)We are now in a position to complete the proof of Theorem 12.6.2. By 12.6.18.2,
Kv is an A 7 -block. Therefore S = Tv, since A6 is of odd index in A1. Hence
Rv = R = Cs(Lv/02(Lv)) = 02(KvS). Since the natural module for A1 has
trivial 1-cohomology by I.1.6.1, 02 (KvS) = UCs(Kv) by C.1.13.b. Then from
the structure of the A1-module, Cs(Lv) = Cs(Kv) x (vo). By 12.6.18.4, LT =
LS, so To := CT(L) n Cs(Kv) :":l LT = LS, and hence To = 1 as Kv f:. Mby 12.6.11. Then as CT(L) :S Cs(Lv), JCT(L)I :S JCs(Lv) : Cs(Kv)I = 2, so
CT(L) = Zv as Zv-:/=-1by12.6.18.3. Therefore as the 1-cohomology of the natural
module for As is 1-dimensional by I.1.6.1, we conclude from C.1.13.b that either
. Q = V, or Q is isomorphic to the 8-dimensional permutation module P as an
L/V-module. In either case Rv = (r)Q with f = (1, 2) and [Rv, r] = [Q, r] = (v).
Also JRvl = 2JQJ = 2s or 29. On the other hand as Rv = Cs(Kv) x U with
U the 6-dimensional permutation module for Kv/02(Kv) ~ A1, r E CRv (Lv) =
(v 0 ) x Cs(Kv)· In particular r centralizes U, so as [Rv,r] = (v), [Cs(Kv),r] = (v).
As Rv is nonabelian, but U is central in in Rv, we conclude Cs(Kv) is nonabelian,
so JCs(Kv)I 2: 8 and hence IRvJ 2: 29. Then using our earlier bounds, IRvl = 2^9 ,
with Q ~ P ~ E2s and Rv ~ Ds X E54. As Rv = 02(KvS) and Kv = 02 (Kv),
Kv acts on both E 2 s-subgroups of Rv, so that Kv :::; Na(Q) :::; M, for our usual
contradiction to 12.6.11.
This contradiction completes the proof of Theorem 12.6.2 ..In the remainder of the subsection, we show Zv = 1.
LEMMA 12.6.19. {1) L controls G-fusion in V.
(2) If Zv-:/=- 1 and V4 is of weight 4 in V, then JCa(v4) : CM(v4)J is odd.
PROOF. Suppose Zv = 1. Then (2) is vacuous, and L has two orbits on V#,
consisting of the singular and nonsingular vectors, with the singular vectors 2-
central. By Theorem 12.6.2, the nonsingular vectors are not 2-central, so (1) holds
in this case.
Thus we may assume Zv -:/=-1. In this case, L has four orbits on V#, withrepresentatives v2, v4, V5, vs, where Vm is of weight m. By Theorem 12.6.2,
JCa(vm)l2 = JTJ/4 form= 2,6. Notice vs is 2-central, and JCM(v4)J2 = ITJ/2.
Assume that (1) fails. Then it follows that vf = vs for some g E G. Wemay choose V4 so that V1 = (v4,vs); thus T4 := CT(v4) E Syb(CM(v4)) and
02 (CL(v4)) = 0^2 (NL(V1)) =Li. Then Lf. :S Ca( vs) :SM, so Lf. :SL by 12.6.1.5,
and we may take Tf :::; T. As JT: T4J = 2 and LiT4/R1 is of index at most 2 in
S3 x S3, we conclude Li E Lf; thus as L centralizes vs, we may take Li = Li. But
then Rf = 02(L1T4)B = Ri. Now using 12.6.1.4, g E Na(R1) :::; M, contrary toV4 rJ. v:1. Hence (1) is established.
Suppose finally that (2) fails. Then vf E Z for some g E G. If [V, J(T)] = 1,
then Na(J(T)) :::; M by 3.2.10,1; and by Burnside's Fusion Lemma, Na(J(T))
controls fusion in Z(J(T)) 2: VZ, contrary to v 4 not 2-central in M. Thus we
may take L = [L, J(T)], so as Zv -:/=-1, L centralizes Z by 12.6.1.6. Hence Lf. :::;
Ca(vf) :::; M = !M(LT). We now repeat the argument of the previous paragraph
to obtain the same contradiction, completing the proof of (2). D
LEMMA 12.6.20. Let S be the set of vectors in V of weight 4. If g E G-Na(V)
with V n VB-:/=-1, then
{1) V n VB s:; s, so m(V n VB) ::::: 3.